Theorem difdif 4065

Description: Double class difference. Exercise 11 of [TakeutiZaring] p. 22. (Contributed by NM, 17-May-1998.)

Assertion

Ref	Expression
difdif	⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴

Proof of Theorem difdif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pm4.45im 825	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)))
2		iman 402	. . . . 5 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) ↔ ¬ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
3		eldif 3897	. . . . 5 ⊢ (𝑥 ∈ (𝐵 ∖ 𝐴) ↔ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴))
4	2, 3	xchbinxr 335	. . . 4 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) ↔ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴))
5	4	anbi2i 623	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)))
6	1, 5	bitr2i 275	. 2 ⊢ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
7	6	difeqri 4059	1 ⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   = wceq 1539   ∈ wcel 2106   ∖ cdif 3884

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-v 3434  df-dif 3890

This theorem is referenced by:  dif0  4306  undifabs  4411